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Unformatted text preview: 22.02 Problem Set 3 Solution 1.Solution The energy eigenvalue problem for the given system is: H = E h 2 d 2 H = 2 m dx 2 + V ( x )  x  > a V ( x ) = V  x  a For bounded particle, the solution is: = Ae x x < a 1 2 = B sin kx or B 'cos kx a x a 3 = Ce x x > a k = 2 0 ) ( 2 h V E m + , = 2 2 h mE Since the solution has to be odd, 2 = B sin kx Also we know, 2 ( a ) = 3 ( a ) , if 3 ( x ) is decaying, there has to be at least a quarter of wave from 0 to a within the well. This means / 4 a . Just being bounded means / 4 = a and E = 0 . This gives us k = m a V mV a 2 2 2 0 2 0 8 2 4 2 2 h h = = = 2.Solution Liboff 5.12 (a) Proof [ A + B , C ] = [ A , C ] + [ B , C ] LHS = [ A + B , C ] = ( A + B ) C C ( A + B ) = A C + B C C A C B RHS = [ A , C ] + [ B , C ] = A C C A + B C C B = LHS (b) Proof [ AB , C ] = A [ B , C ] + [ A , C ] B LHS = ABC CAB RHS = A ( B C C B ) + ( A C C A ) B = A B C A C B + A C B C A B = A B C C A B = LHS Liboff 5.19 ( 2 , 3 ) = {sin kx ,exp( ikx )} Since we know 3 = exp( ikx ) is a common eigenfunction of H and p , wed like to keep it as one of the two common eigenfunctions composed of ( 2 , 3 ) . If we denote it as a , then a = a 2 2 + a 3 3 = 3 , which means a 2 = 0 and a 3 = 1 . (1) Also we know b = exp( ikx ) is a common eigenfunction of H and p and is linearly independent of a = exp( ikx ) , we hope we could construct b as a linear combination of ( 2 , 3 ) . b = b 2 2 + b 3 3 = b 2 sin kx + b 3 exp( ikx ) = b 2 sin kx + b 3 (cos kx + i sin kx ) 22.02 Problem Set 3 Solution b = b 3 cos kx + ( b 2 + ib 3 )sin kx = exp( ikx ) = cos kx i sin kx b 3 = 1, b 2 = 2 i (2) Sum (1) and (2), we get: a = 0 2 + 1 3 = exp( ikx ) b = 2 i 2 + 1 3 = exp( ikx ) 2 and 3 are a pair of linearly independent common eigenfunctions of H and p . Liboff 5.25 Use uncertainty principle: If [ A , B ] = C 0 , then A B 1  < C >  2 (a) x E [ x , E ] = x E E x h 2 d 2 x = x , E = 2 m dx 2 + V ( x ) h 2 d 2 h 2 d 2 [ x , E ] = x [ 2 m dx 2 + V ( x ) ] [ 2 m dx 2 + V ( x )][ x ] h 2 d 2 h 2 d 2 = x 2 m dx 2 + xV ( x ) + 2 m dx 2 [ x ] xV ( x ) h 2 d 2 h 2 d d h 2 d 2 h 2 d h 2 d 2 h 2 d + + + x + = x 2 m dx 2 2 m dx [ + x dx ] = x 2 m dx 2 2 m dx 2 m dx 2 2 m dx [ x , E ] = h 2 d m dx d p x = i h dx [ x , E ] = i h p x m 1 h From uncertainty principle, we get x...
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 Spring '09
 Mucciolo
 mechanics, Energy

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